114/6=9 parts per minute. 9 parts * 15 minutes= 135 parts
Step-by-step explanation:
a) 7!
In absence of any restrictions, the answer is 7! as it represents the permutations of all animals.
b) 4! x 3!
Considering there are 6 cats and 5 dogs, the first and last slots must be occupied by cats to ensure alternate arrangements. The only options available then are based on the arrangement of the cats among themselves and the dogs among themselves, yielding 4! permutations for the cats and 3! for the dogs, thus leading to a total of 4! x 3! arrangements.
c) 3! x 5!
Here, the arrangement of the dogs among themselves can occur in 3! ways. Considering the dogs as a singular “object,” we can arrange this unit with the 4 cats, providing 5! total arrangements possible, leading to 3! · 5! arrangement possibilities.
d) 2 x 4! x 3!
In this scenario, both cats and dogs must be grouped together, allowing positions where all cats come before the dogs or vice versa. As there are two configurations, the resultant count is 2 multiplied by both arrangements, resulting in 2 x 4! x 3!
Among the options provided, the correct choice is: D 7y^4-13x^3 inches
Point Q'Q ′ Q, prime is the image of Q(-7,-6)Q(−7,−6)Q, left parenthesis, minus, 7, comma, minus, 6, right parenthesis under the
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Answer:
5,2
Step-by-step explanation: Adding -7 and 12 results in 5, while for -6 it gives 2 under the equation (8=2) and I confirmed this on Khan Academy!:)
I hope this is helpful to you!!
Only amend the first item. Please refer to the graph in the attached file. 1) a rectangle with A(3,3), B(3,6), C(7,6), D(7,3) having adjacent sides (in green) 2) a parallelogram with A(2,0), B(3,2) having nonperpendicular sides (in blue) C(6,3), D(5,1) 3) a square with A(3,3), B(2,5), C(4,6), D(5,4) (in red) 4) a rhombus with A(2,-2), B(3,0), C(4,-2), D(3,-4) having adjacent sides (in black)
